Ask question

Ask question

All categories

2 years ago

7

A .100 kg bullet is flying at 200 m/s. How much energy is the bullet storing due to its motion?

1 answer:

morpeh [17]2 years ago

4 0

KE = 2000 J

Explanation:

KE = (1/2)mv^2

= (1/2)(0.100 kg)(200 m/s)^2

= 2000 J

You might be interested in

Assapp!!!’!!!!!!!! !!!!!!

dlinn [17]

Answer:

delta r(x) = (delta (r)) * cos(alpha), delta r(y) = (delta(r)) * sin(alpha)

Explanation:

Well it's a simple rule I guess...

6 0

3 years ago

A wave pulse travels down a slinky. The mass of the slinky is m = 0.87 kg and is initially stretched to a length L = 6.8 m. The

Ber [7]

Answer:

1. $v=14.2259\ m.s^{-1}$

2. $F_T=25.8924\ N$

3. $\lambda=29.6373\ m$

Explanation:

Given:

mass of slinky, $m=0.87\ kg$

length of slinky, $L=6.8\ m$

amplitude of wave pulse, $A=0.23\ m$

time taken by the wave pulse to travel down the length, $t=0.478\ s$

frequency of wave pulse, $f=0.48\ Hz=0.48\ s^{-1}$

1.

$\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel$

$v=\frac{6.8}{0.478}$

$v=14.2259\ m.s^{-1}$

2.

<em>Now, we find the linear mass density of the slinky.</em>

$\mu=\frac{m}{L}$

$\mu=\frac{0.87}{6.8}\ kg.m^{-1}$

We have the relation involving the tension force as:

$v=\sqrt{\frac{F_T}{\mu} }$

$14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }$

$202.3774=F_T\times \frac{6.8}{0.87}$

$F_T=25.8924\ N$

3.

We have the relation for wavelength as:

$\lambda=\frac{v}{f}$

$\lambda=\frac{14.2259}{0.48}$

$\lambda=29.6373\ m$

8 0

2 years ago

In which way are the cathode rays deflected in the x-plates of the Cathode Ray Oscilloscope

MrRissso [65]

Answer:

The cathode ray is deflected vertically to the fluorescent screen

Explanation:

$.$

4 0

2 years ago

Read 2 more answers

What would be the current going through a 200 ohm resister that is connected across a 120 v power supply?

gavmur [86]

Here is your answer

hope it help you

6 0

3 years ago

The speed of a certain proton is 350 km/s. If the uncertainty in its momentum is 0.100%, what is the necessary uncertainty in it

Evgesh-ka [11]

Answer:

$\Delta x = 1.807 \times 10^{-10}m$

Explanation:

mass of proton, m = 1.67 x 10^-27 kg

speed of proton, v = 350 km/s = 350,000 m/s

Momentum of proton, p = mass x speed

p = 1.67 x 10^-27 x 350000 = 5.845 x 10^-22 kg m /s

uncertainty in momentum, Δp = 0.1 % of p

Δp = $\frac{0.1\times 5.845 \times 10^{-22}}{100}=5.845 \times 10^{-25}$

According to the principle

$\Delta x\times \Delta p \geq \frac{h}{2\pi }$

where, Δx be the uncertainty in position

$\Delta x\times 5.845 \times 10^{-25}= \frac{6.634 \times 10^{-34}}{2\times 3.14}$

$\Delta x = 1.807 \times 10^{-10}m$

5 0

2 years ago